What is the formula for calculating the FWHM of a gaussian function?

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The discussion focuses on calculating the Full Width at Half Maximum (FWHM) for a Gaussian function defined as exp[-(π*x²/A²)]. The FWHM is established as 0.939A, but the proof is sought. The method involves setting the Gaussian equal to 0.5 to find the half-maximum and applying natural logarithms to simplify the equation. The variable substitution x = kA is introduced, leading to the computation of k as (-ln(0.5)/π)^(1/2). The conversation highlights the challenge in proving the FWHM formula for this specific Gaussian function.
peterjaybee
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Hi,

I have a gaussian of the form
exp[-\frac{\pi*x^{2}}{A^2}].

I know that the FWHM=0.939A, but I cannot prove it.

I Let exp[-\frac{\pi*x^{2}}{A^2}=0.5 (i.e. the half maximum part)

taking natural logs I get rid of the exponential, but then which bit represents the full width?
 
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Write x = kA and compute k=(-ln(.5)/π)1/2

Your Gaussian looks funny.
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

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